$a^n-a + 1 $ divisible by $n$
$begingroup$
Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?
number-theory elementary-number-theory divisibility
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked Oct 24 '18 at 8:10
DronaDrona
1627
$endgroup$
|
show 12 more comments
$begingroup$
Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?
number-theory elementary-number-theory divisibility
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked Oct 24 '18 at 8:10
DronaDrona
1627
$endgroup$
5
$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30
3
$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48
1
$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16
2
$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54
6
$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39
|
show 12 more comments
$begingroup$
Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?
number-theory elementary-number-theory divisibility
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked Oct 24 '18 at 8:10
DronaDrona
1627
$endgroup$
Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?
number-theory elementary-number-theory divisibility
number-theory elementary-number-theory divisibility
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked Oct 24 '18 at 8:10
DronaDrona
1627
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked Oct 24 '18 at 8:10
DronaDrona
1627
edited Dec 7 '18 at 16:57
Klangen
1,73711334
edited Dec 7 '18 at 16:57
Klangen
1,73711334
edited Dec 7 '18 at 16:57
Klangen
1,73711334
1,73711334
asked Oct 24 '18 at 8:10
DronaDrona
1627
asked Oct 24 '18 at 8:10
DronaDrona
1627
asked Oct 24 '18 at 8:10
DronaDrona
1627
1627
5
$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30
3
$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48
1
$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16
2
$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54
6
$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39
|
show 12 more comments
5
$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30
3
$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48
1
$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16
2
$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54
6
$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39
5
5
$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30
$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30
3
3
$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48
$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48
1
1
$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16
$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16
2
2
$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54
$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54
6
6
$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39
$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39
|
show 12 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$N=a^n-a+1$
$a^{p_1} ≡ a mod p_1$
$a^{p_2 } ≡a mod p_2$
$(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$
$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$
⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$
If $n=p_1p_2 | a^n-a+1$ then we must have:
$p_1p_2 | k_1p_1 + k_2p_2+1$
So we have following linear equation:
$k_1p_1 + k_2 p_2 = m p_1p_2-1$
For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:
with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:
$k_1= 7 t + 11$ and $k_2= -5 t+7$.
Now in first step problem reduces to:
Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.
For example for $a=3$, $p_1=5$ and $p_2=7$ we have:
$3^{35}-3+1=105$ and $(35, 105)=5$
In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.
Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:
$k_1p_1+k_2p_2= M(a)$
Or: $a | k_1p_1+k_2p_2$
We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$
This can help us in choosing a and primes $p_1$ and $p_2$
I see no reason for the lack of more solutions.
answered Feb 13 at 9:19
siroussirous
1,6851514
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2968838%2fan-a-1-divisible-by-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$N=a^n-a+1$
$a^{p_1} ≡ a mod p_1$
$a^{p_2 } ≡a mod p_2$
$(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$
$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$
⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$
If $n=p_1p_2 | a^n-a+1$ then we must have:
$p_1p_2 | k_1p_1 + k_2p_2+1$
So we have following linear equation:
$k_1p_1 + k_2 p_2 = m p_1p_2-1$
For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:
with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:
$k_1= 7 t + 11$ and $k_2= -5 t+7$.
Now in first step problem reduces to:
Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.
For example for $a=3$, $p_1=5$ and $p_2=7$ we have:
$3^{35}-3+1=105$ and $(35, 105)=5$
In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.
Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:
$k_1p_1+k_2p_2= M(a)$
Or: $a | k_1p_1+k_2p_2$
We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$
This can help us in choosing a and primes $p_1$ and $p_2$
I see no reason for the lack of more solutions.
answered Feb 13 at 9:19
siroussirous
1,6851514
$endgroup$
add a comment |
$begingroup$
$N=a^n-a+1$
$a^{p_1} ≡ a mod p_1$
$a^{p_2 } ≡a mod p_2$
$(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$
$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$
⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$
If $n=p_1p_2 | a^n-a+1$ then we must have:
$p_1p_2 | k_1p_1 + k_2p_2+1$
So we have following linear equation:
$k_1p_1 + k_2 p_2 = m p_1p_2-1$
For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:
with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:
$k_1= 7 t + 11$ and $k_2= -5 t+7$.
Now in first step problem reduces to:
Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.
For example for $a=3$, $p_1=5$ and $p_2=7$ we have:
$3^{35}-3+1=105$ and $(35, 105)=5$
In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.
Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:
$k_1p_1+k_2p_2= M(a)$
Or: $a | k_1p_1+k_2p_2$
We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$
This can help us in choosing a and primes $p_1$ and $p_2$
I see no reason for the lack of more solutions.
answered Feb 13 at 9:19
siroussirous
1,6851514
$endgroup$
add a comment |
$begingroup$
$N=a^n-a+1$
$a^{p_1} ≡ a mod p_1$
$a^{p_2 } ≡a mod p_2$
$(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$
$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$
⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$
If $n=p_1p_2 | a^n-a+1$ then we must have:
$p_1p_2 | k_1p_1 + k_2p_2+1$
So we have following linear equation:
$k_1p_1 + k_2 p_2 = m p_1p_2-1$
For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:
with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:
$k_1= 7 t + 11$ and $k_2= -5 t+7$.
Now in first step problem reduces to:
Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.
For example for $a=3$, $p_1=5$ and $p_2=7$ we have:
$3^{35}-3+1=105$ and $(35, 105)=5$
In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.
Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:
$k_1p_1+k_2p_2= M(a)$
Or: $a | k_1p_1+k_2p_2$
We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$
This can help us in choosing a and primes $p_1$ and $p_2$
I see no reason for the lack of more solutions.
answered Feb 13 at 9:19
siroussirous
1,6851514
$endgroup$
$N=a^n-a+1$
$a^{p_1} ≡ a mod p_1$
$a^{p_2 } ≡a mod p_2$
$(a^{p_1})^{p_2} ≡ a^{p_2} mod p_1 ≡(a mod p_2) mod p_1= k_1 p_1 + k_2 p_2 + a$
$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$
⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$
If $n=p_1p_2 | a^n-a+1$ then we must have:
$p_1p_2 | k_1p_1 + k_2p_2+1$
So we have following linear equation:
$k_1p_1 + k_2 p_2 = m p_1p_2-1$
For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:
with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:
$k_1= 7 t + 11$ and $k_2= -5 t+7$.
Now in first step problem reduces to:
Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.
For example for $a=3$, $p_1=5$ and $p_2=7$ we have:
$3^{35}-3+1=105$ and $(35, 105)=5$
In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.
Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1mod a$ and $p_2≡1mod a$, then:
$k_1p_1+k_2p_2= M(a)$
Or: $a | k_1p_1+k_2p_2$
We can see this in solution $n=409times 9831853$ for $a=6$; $409=48times 6 +1$ and $9831853=1638642 times 6 +1$
This can help us in choosing a and primes $p_1$ and $p_2$
I see no reason for the lack of more solutions.
answered Feb 13 at 9:19
siroussirous
1,6851514
answered Feb 13 at 9:19
siroussirous
1,6851514
answered Feb 13 at 9:19
siroussirous
1,6851514
answered Feb 13 at 9:19
siroussirous
1,6851514
1,6851514
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2968838%2fan-a-1-divisible-by-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
5
$begingroup$
Interesting question. Have you tried anything yet?
$endgroup$
– Mohammad Zuhair Khan
Oct 24 '18 at 8:30
3
$begingroup$
The case $a=2$ is popular, see here and the related links.
$endgroup$
– Dietrich Burde
Oct 24 '18 at 8:48
1
$begingroup$
$n$ is not a prime number. It also must be odd.
$endgroup$
– Oldboy
Oct 24 '18 at 18:16
2
$begingroup$
My original problem: Let $ainmathbb Z$ and $a>3$, prove that there exist infinitely many positive integers $n$ satisfying$$(n+a)mid left(a^n+1right).$$
$endgroup$
– Drona
Oct 26 '18 at 19:54
6
$begingroup$
@Drona I think that you are making a big mistake. The statement $(n+a)|(a^n+1)$ is not equivalent to $n|(a^n-a+1)$. It's like saying that $(4+3)|14$ is equivalent to $4|(14-3)$. You should post the original problem. If you don't want to do it, I would like to do it.
$endgroup$
– Oldboy
Oct 27 '18 at 8:39